Do you know the correct answer?

[miketaylor]

I apologise for not having read everything since I originally posted, but am somewhat short of time.

However, respondents seem to have missed my point. Perhaps I was not clear enough.

Contestant chooses a door. Monty opens one of the two remaining doors to reveal a goat.

Statistics say that Contestant can double his chances of winning the car by switching.

But it doesn't seem to matter which door the contestant originally chose.

He chose door 1; his chances double by switching to door 2.

He chose door 2; his chances double by switching to door 1.

This is the peculiarity i would like to see explained.

[Tom Wilkinson]

I apologise for not having read everything since I originally posted, but am somewhat short of time.

However, respondents seem to have missed my point. Perhaps I was not clear enough.

Contestant chooses a door. Monty opens one of the two remaining doors to reveal a goat.

Statistics say that Contestant can double his chances of winning the car by switching.

But it doesn't seem to matter which door the contestant originally chose.

He chose door 1; his chances double by switching to door 2.

He chose door 2; his chances double by switching to door 1.

This is the peculiarity i would like to see explained.

Because all doors are equal. They all have an equal chance to win.

Once you have chosen a door, you have a 1 in three chance of winning. Monty has a 2 in 3 chance. You know one of his is wrong from the start. Your choice of door doesn’t affect the odds. And he can always show you one door with a goat because he HAS to have one, and he knows which it is.

[oldcodger]

Simple explanation

[oldcodger]

Comprehensive lecture by a PhD at Harvard, first 25 minutes of his talk.



Edited to add, the lecturer is Joe Blitzstein, Professor of the Practice in Statistics, Harvard University, Department of Statistics

[lupussonic]

What seems to be the consensus amongst some here is that the set of probabilities of first choice (1/3) is carried over somehow to the second set of probabilities in the second choice (1/2).

Nick (and I) can not see that they are connected in anyway. How are the 2 sets of probabilities connected?

To my mind it doesn't matter a damn which door I chose on first choice, or which door is the goat reveal, the second choice of 2 doors is completely unrelated to the first choice of 3 doors, yet some here say they are entangled some how?

So what's the mechanism?

[Tom Wilkinson]

What seems to be the consensus amongst some here is that the set of probabilities of first choice (1/3) is carried over somehow to the second set of probabilities in the second choice (1/2).

Nick (and I) can not see that they are connected in anyway. How are the 2 sets of probabilities connected?

To my mind it doesn't matter a damn which door I chose on first choice, or which door is the goat reveal, the second choice of 2 doors is completely unrelated to the first choice of 3 doors, yet some here say they are entangled some how?

So what's the mechanism?

The are connected because the “game” was never reset. Monty didn’t open a door randomly leaving you with new choices. You new choice is really do I stick with my one door, or do I choose both doors that Monty had. 796 is a good explanation.

What information did you really gain by him showing you what was behind one of his doors? If, before opening a door to reveal a goat Monty asked if you wanted his two doors or stick with your 1 door would you stay or switch? You would switch, because you now have two chances. You know one of them has a goat but the odds of a win are better.

[ShorelineJohn]

I want an equation, I know it has to exists. else the answer is wrong!
And 1 car 1 medium prize and 1 goat, not 2 goats!

[Tom Wilkinson]

I want an equation, I know it has to exists. else the answer is wrong!
And 1 car 1 medium prize and 1 goat, not 2 goats!

The Harvard vid shows the equation, around 14 min mark. Worth watching.

[Tom Montgomery]

I want an equation, I know it has to exists. else the answer is wrong!

OK. Go here: https://medium.com/@judaikawa/the-mo...m-4415e50e233f. The article to which I've linked shows a different statistical equation solving the solution than that in the Harvard video (using the Law of Total Probability). The Harvard video is very good. It is also clear (I could follow it).

After viewing the video and then reading the linked article and working through the Bayesian equation, actually play the game yourself. Do what JimmyW suggested, take an Ace and two Jokers out of a deck of playing cards. Enlist a friend to take the Monty Hall role. Play over and over for an hour. Bet a dollar to win on each play. Never switch. Be sure to return and tell us if your friend thanked you.
.

[AnalogKid]

What seems to be the consensus amongst some here is that the set of probabilities of first choice (1/3) is carried over somehow to the second set of probabilities in the second choice (1/2).

Nick (and I) can not see that they are connected in anyway. How are the 2 sets of probabilities connected?

To my mind it doesn't matter a damn which door I chose on first choice, or which door is the goat reveal, the second choice of 2 doors is completely unrelated to the first choice of 3 doors, yet some here say they are entangled some how?

So what's the mechanism?

You accept that the initial choice is a 1/3 change of getting the car.

If you choose to switch your first choice for Monty's other door, then you reverse your choice - if you chose the car and switch, you get a goat, if you chose a goat and switch you get a car.

Therefore, unless you can come up with a scenario where switching to Monty's unrevealed door does not reverse the outcome of your initial choice, switching doors must have a 2/3 chance of winning the car.

[Tom Montgomery]

.
The offer is to switch your one door (with a 1/3 chance of having the car) for the aggregate of the remaining doors (which have a 2/3 chance of having the car). The odds the aggregate has a 2/3 chance of having the car do not change if one of the doors of the aggregate is revealed to not contain the car. You knew that already.

[Tom Montgomery]

.
It is a simple game to play with cards representing the doors. THEY WILL NOT PLAY.

[Keith Wilson]

the second choice of 2 doors is completely unrelated to the first choice of 3 doors, yet some here say they are entangled some how?

After you pick, Monty has two doors. At least one of them will always hide a goat, and Monty know what's behind them. He away opens one with a goat, never the car. Thus, if you switch, you're getting the best of the two doors.

Again, a proposition: We have two jokers and an ace. We shuffle and deal you one card, two for me. I'll always have a joker. I look at my cards, and show you a joker. We then turn over our remaining cards, and the ace wins. I'll play you for ten bucks a hand, for as long as you like. Try it yourself, it won't take long.

[Tom Montgomery]

.
It is so simple. I am baffled why not one of the skeptics will play the game for themselves. Maybe because once observation contradicts one's hypothesis the hypothesis must be abandoned. I played the game 100 times. Had the total result been anywhere near 50/50 I would have admitted being in error.

[Tom Montgomery]

.
I received a pm objecting that it is impossible to shuffle three cards and always have a random distribution.

OK. Then play the game using three marbles rather than cards, say one black and two white. Play with a friend. Place the marbles in a bag, have your friend shake it well, and then you draw one marble from the bag. You must hold it in your clenched fist and not look at it. Your friend gets the bag and he may look inside to see what marbles he has. He may take out one white marble, show it to you, and then put it to one side. He then offers to switch the marble in his bag with the one in your hand. Turn down the offer. Reveal the marbles. Do this a number of times. Better?

The other objection was that the game must be played hundreds of times to prove anything... and that life is too short. So probably not better.
.

[coelcanth]

What seems to be the consensus amongst some here is that the set of probabilities of first choice (1/3) is carried over somehow to the second set of probabilities in the second choice (1/2).

Nick (and I) can not see that they are connected in anyway. How are the 2 sets of probabilities connected?

To my mind it doesn't matter a damn which door I chose on first choice, or which door is the goat reveal, the second choice of 2 doors is completely unrelated to the first choice of 3 doors, yet some here say they are entangled some how?

So what's the mechanism?

the mechanism is that all three doors are in the same game pool of three, and selecting the first of them removes it from that pool.
it is a layer of probabilities in which the first pick affects the second pick in the game.
even without agreeing on the exact number of probability, you cannot assert that the first pick has no bearing on the outcome of the game.

we must accept that the player always has a 1 in 3 chance of winning the prize on their first guess.
if Monty were playing against us, yes, his pick would only be from two doors.. but we could not say it is a 1 in 2 chance because that would not account for the 1 in 3 times the contestant might have won right from the start and Monty might, in fact, be holding two goats to choose from.

at the very beginning of the game, there are three doors all with equal chance of hiding the prize (1/3 + 1/3 + 1/3)
at the moment the player chooses their first door, the pool is split into two unequal parts: <- this is the big event and the mechanism that knocks the earth off its axis

one player's door (with 1/3 chance of hiding the prize)
and Monty's two doors (one door with 1/3 chance and the other door with equal 1/3 chance) Monty's probability 2/3 = 1/3 + 1/3

there is never a 1 in 2 chance that either of Monty's doors has the prize, because that would not account for the first player's door which was loaded at the same time as Monty's
the only 1 in 2 chance is whether Monty will open his left door or his right door, and that does not matter.

even Nick, i think, would agree that at the point in the game when the first door was picked (but no doors were opened) the chances are 1/3 v. 2/3
so what if, at this point in the game, Monty didn't offer his 'deal' but we all simply agreed to open all three doors at the exact same time ?
what would happen is, player's door would win 1 in 3 times and Monty's two doors together as a whole would win 2 in 3 times.. there's no way around that

when Monty decides to reveal a door and offer a deal before the final winner is decided, he does not alter the math in any way. he has just thrown in a bit of human psychology.

lets think about it rationally..
before the reveal, player's door has a 1/3 chance of winning the prize and Monty's two doors a total 2/3 chance
we also know (without seeing) that one of Monty's doors must contain one goat. we know this because player's door can only hold one goat or the prize at one time, but there are two goats in the game.
at this point (take a deep breath) Monty gingerly opens a door and reveals one of the game's two goats. this revelation only tells us specifically which door the goat is behind. but we are not actually interested in this because the game is not asking us to pick which of Monty's two doors has the prize. in the rules of the game, the final choice is from one of Monty's doors or the only one of the Player's door.
the important part is, by opening one of his doors, Monty has reduced our choice from two down to the last one of his doors but not actually changed the math at all. that's because the total probability that his two doors contain the prize is still 2 in 3, even though we now only have one of his doors to choose from.

[Tom Montgomery]

.
One's chances of choosing the one door of three containing a car is NEVER a 50% probability. It is ALWAYS a 33.33% probability. The skeptics would have us believe the way the game operates changes the odds off that initial choice. That is impossible.

[Keith Wilson]

If Monty initially opened one door to reveal a goat, then asked you to pick between the remaining two, the odds would indeed be 50-50. That is NOT the same thing, although the game is designed to fool you into thinking it is.

[George.]

PM,

I understand your position. I really do. For the first 14 pages of this thread I only lurked, because I intuitively thought like you - that it has to be 50-50 between two doors, and the opening of the goat door cannot affect the odds.

But I also knew that this is wrong, not because I understood it, but because everyone, including unimpeachable sources, told me so. So I had to conclude that my logic was flawed, even if it seemed perfect to me.

Only around page 14 (I actually only skimmed a few of the preceding posts) I took a walk with the dogs and tried to think about the problem with a different approach, and suddenly I understood that Monty has two doors. The one he opens is deliberately chosen to not reduce his odds of winning. Anyway, it doesn't matter how I saw the light. What matters is that you need to take fresh approaches to the problem until one of them gives you that aha! moment and you too see the light.

[George Jung]

.
I received a pm objecting that it is impossible to shuffle three cards and always have a random distribution.

OK. Then play the game using three marbles rather than cards, say one black and two white. Play with a friend. Place the marbles in a bag, have your friend shake it well, and then you draw one marble from the bag. You must hold it in your clenched fist and not look at it. Your friend gets the bag and he may look inside to see what marbles he has. He may take out one white marble, show it to you, and then put it to one side. He then offers to switch the marble in his bag with the one in your hand. Turn down the offer. Reveal the marbles. Do this a number of times. Better?

The other objection was that the game must be played hundreds of times to prove anything... and that life is too short. So probably not better.
.

Ha! I like that. It's much better to waste ones life arguing about it for days on end!

[peb]

If Monty initially opened one door to reveal a goat, then asked you to pick between the remaining two, the odds would indeed be 50-50. That is NOT the same thing, although the game is designed to fool you into thinking it is.

This is most easily seen with a deck of cards. If you are hoping for the ace of spades, you have 1/52 chance. If Monty randomly picks a card, his is 1/52. Now if you start turning over the other 50, 1 out of 26 times you won't find the ace of spades. So it is 50-50 between you and Monty. OTOH, if Monty goes through and specifically picks out the ACE of spades, leaves it turned over and then you go thought the other 50 cards, at that point Monty has a 51/52 chance of winning. Perhaps the thing that confuses people is stating the rule that Monty has to open a door with a goat. Instead look at it, he has to leave the car hidden.

[Jimmy W]

.
I received a pm objecting that it is impossible to shuffle three cards and always have a random distribution.

OK. Then play the game using three marbles rather than cards, say one black and two white. Play with a friend. Place the marbles in a bag, have your friend shake it well, and then you draw one marble from the bag. You must hold it in your clenched fist and not look at it. Your friend gets the bag and he may look inside to see what marbles he has. He may take out one white marble, show it to you, and then put it to one side. He then offers to switch the marble in his bag with the one in your hand. Turn down the offer. Reveal the marbles. Do this a number of times. Better?

The other objection was that the game must be played hundreds of times to prove anything... and that life is too short. So probably not better.
.

Ha! I like that. It's much better to waste ones life arguing about it for days on end!

If played for $10 a hand, one will probably decide to quit giving away money long before 100s of times.

[Breakaway]

How can the probabilities of choosing between 2 different doors be 30% and 50%?

The choice is between three doors. Two doors is just a stop enroute to the destination( end of game).


Sent from my iPhone using Tapatalk Pro

[Decourcy]

Exposing a card or door seems to be the mental hurdle. So just forget that and think about the odds with them hidden. One person has 2, one person has 1. Who would ever bet that the person with 1 has better odds of having the prize….?

[Tom Wilkinson]

Exposing a card or door seems to be the mental hurdle. So just forget that and think about the odds with them hidden. One person has 2, one person has 1. Who would ever bet that the person with 1 has better odds of having the prize….?

Apparently a couple people on the WBF.

[David G]

Tom - you are a genius.

I know... I know... you claim it was an accident.

But that just makes you and Accidental Genius!!

[Dave Wright]

Exposing a card or door seems to be the mental hurdle. So just forget that and think about the odds with them hidden. One person has 2, one person has 1. Who would ever bet that the person with 1 has better odds of having the prize….?

Maybe this situation has reached the "Trumpian phase," where the search for a cover up or face saving "out" is taking all the mental effort, and the truth of the problem has been abandoned...

[Tom Montgomery]

^
No. I believe those who believe the choice comes down to a 50/50 proposition are sincere in that belief. Any error must be in the assumptions behind their calculation.

If it were me, and most people were telling me I was wrong, I would test my belief with a real-world test. That is in fact what happened with me. When I first encountered the problem I thought there was no advantage to switching. But upon investigating further I found an overwhelming number of sources providing logical explanation and statistical calculation that changed my mind.

Nevertheless the solution to the problem certainly is counter-intuitive. Actually playing the game 100 times eliminated any remaining doubt I had regarding the advantage in switching. The critics claim that my results were due to being unable to randomly shuffle only three cards. I would play it again using three marbles. But I do not have marbles. Are they even commonly found for sale these days? In any event I suspect some other objection would discount the result of that form of the game as well.
.

[birlinn]

I think the problem results from not realising that the host knows what is behind all three doors- once the goat is revealed, the contestant knows only two.

[birlinn]

I think the problem results from not realising that the host knows what is behind all three doors- once the goat is revealed, the contestant knows only two possibilities..

[Gordon Bartlett]

Wow…. 20 pages.

I’ve been out for the last half dozen pages, so I’ll take a swing.

No one reads my sh!t anyways.

Nick,

If Monty was selecting from his two doors at random and happened to select a goat, then you would be correct, the odds of either door containing the car would be 50/50.

But he’s not. Monty knows the locations of the car and goats from the outset. The draw is only random for the contestant.

Once Monty makes the informed choice to remove a goat from his two, the remaining door now has the odds are locked in a 1/3 and 2/3.

This is 100% correct.
If Monty randomly flung open either of his two doors, the two remaining doors would have identical odds. That would be 50/50 if the door revealed a goat, and 0/0 if the door revealed the car.
But there is nothing random about Monty's choice of door. If the car is to the right, he opens the left. If the car is to the left, he opens the right. The only time he can pick a door randomly is if the contestant has picked the car.
The whole concept of the "always switch" argument revolves around this introduction of Monty's non-random choices to the mix.
If everything remained completely random, there would be no advantage in switching.

[Dave Wright]

I don't believe that it is counterintuitive.

You can put a gold coin underneath one of three identical cups. You switch them around a bit.

Then you ask a child, " Do you think you can guess which cup the coin is under." The child usually answers yes, if I'm lucky!"

You then ask the child, "If I let you turn over two cups rather than just one do think you might have a better chance of finding the coin?

A normal child gets the correct answer very quickly.

The Monte Hall depiction of this problem is full of goats, doors, and other confusing nonsense. Some people are bamboozled with bullsh!t. The real solution is intuitive when the problem is presented simply and unadorned.

[Tom Wilkinson]

^
No. I believe those who believe the choice comes down to a 50/50 proposition are sincere in that belief. Any error must be in the assumptions behind their calculation.

If it were me, and most people were telling me I was wrong, I would test my belief with a real-world test. That is in fact what happened with me. When I first encountered the problem I thought there was no advantage to switching. But upon investigating further I found an overwhelming number of sources providing logical explanation and statistical calculation that changed my mind.

Nevertheless the solution to the problem certainly is counter-intuitive. Actually playing the game 100 times eliminated any remaining doubt I had regarding the advantage in switching. The critics claim that my results were due to being unable to randomly shuffle a deck of cards. I would play it again using three marbles. But I do not have marbles. Are they even commonly found for sale these days? In any event I suspect some other objection would discount the result of that form of the game as well.
.

So you’ve lost your marbles. No wonder after this thread.

I agree with Dave Wrights assessment above (825). The distractions are throwing people off as they are intended to do.

[Gordon Bartlett]

Too many people are ignoring the difference between a randomly opened door vs a selectively opened door. That's the key to the whole thing.
Randomly opened door = even odds.
Selectively opened door = 66%/33%.

[SchoonerRat]

Originally Posted by coelcanth

no that's not it

as long as the contestant has to pick first, their door will always have worse odds than Monty's two doors

Monty is not cheating at the game and he's not 'removing' the goat from play
even with a 1 in 3 chance and without switching doors, the player will still win the prize sometimes, there's just a smaller chance
it's never 50/50

"It's never 50/50"?????????

For God's sake, man! I just tried to find 1 winner from a pool of 50 possible choices. What are my odds of success? I give you that my odds are not now 50/50. What do you think they are? 1/3, 1/5, 1/10. I have 4 years of undergraduate study in Math. Somehow, during that time I was able to convince the powers that be in the school of letters and science that I had earned a bachelor's degree in mathematics. Even ignoring the 3 or 4 semesters I spent doing grad work in math, I think I've earned enough cred to believably posit that my odds are 1/50. Can we agree on this? I can't clearly hear your answer over the internet, so just nod or shake your head.

Okay, now my new best friend Monty has now given me a second chance to choose a different door. I say that this is a completely new choice that is totally unrelated to the first...except that both choices involve goats and car. I now have new info on which to base my decision. Monty now has 1 door that either does or does not hide my shiny new toy. I have 1 door which either does or does not contain a goat that I wouldn't know what to do with if I won it. How in the world can you look at this new choice and still say that it is not now 50/50. There is no place, other than my door or Monty's door where my prize can possibly exist.

NO PLACE!

I sincerely apologize that the hollow bit of bone which resides upon my cervical vertebrae does not contain sufficient gray matter to more plainly explain my position on this. I'll do my best to sum it up.

I chose door #42 using the well-known mathematical principal of "Eenie Meenie Meiney Moe". The odds of my success are 1/50.

Mr. Python has now given me a new choice. I now know, without a doubt, that either my door, or door #13 is correct. When I originally chose #42, #13 was no more than Eenie to my Moe. Odds for #42, 1/50. Odds for #13, 1/50. But I now have more info that gives me a 50/50 chance of being correct no matter whether or not I choose to go with my first gut feeling or to consider changing my mind. There has been nothing revealed to me in the new info available to me that would lead me to believe that my odds are now any better than 1/50.

I now have a 50/50 choice before me. One of those 1/2 choices has no better than a 1/50 chance of being the right choice. The chances for #42 seem to be less than 1/2. I think this makes the chances for #13 somewhat better than 1/2.

Well, after all of this pontificating, I have to admit that your are right. "It's never 50/50. Changing your mind is better than 50/50. No matter which way you go, there's always a good chance that you're gonna lose. If you accept Monty's offer to change your mind the odds will be in your favor. Every time you play. Without Fail. You might lose, but rest assured, you made the correct choice.